求不定积分∫[1/(u²-2u)]du∫[1/(u²-2u)]du=∫[1/u(u-2)]du=(1/2)∫[1/(u-2)-1/u]du=(1/2)[∫d(u-2)/(u-2)-∫du/u]=(1/2)[ln(u-2)-lnu]+C=ln√[(u-2)/u)]+C求不定积分∫[(1/u²)-2u]du∫[(1/u²)-...